I am learning about operators and some other stuff in linear algebra, and I have never encountered a problem that involves a linear operator which is defined recursively, that is:
$$A: V \to V$$
$$A = \begin{cases} f(A(...)) \text{ in general} \\ \text{<constant>} \text{ if <boundary case>} \end{cases}$$
Matrix determinant is close (if you define it recursively as a sum of products), but $\det \mu A \ne \mu \det A$, so it's not a linear operator. Unless, perhaps, the matrix is populated by members of $\mathbb{Z}_2 = \{0, 1\}$.
Do you know about any interesting linear maps between vector spaces which are defined as recurrent relations?
There is, for example, a recursion for these matrices here: $$ \begin{pmatrix} 1 & 1\\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n\\ F_{n} & F_{n-1} \end{pmatrix},$$
where the Fibonacci numbers are recursively defined by
$$\begin{eqnarray} F_0&=&0\\ F_1&=&1\\ F_{n+1}&=&F_n+F_{n-1} \end{eqnarray}$$